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If $x=2\sin^{2}\alpha - \cos2\alpha, then \:\: x \:\: lies \:\: in\:\: the \:\: interval$

Option 1)
$\left [-1, 3 \right ]$

Option 2)
$\left [1, 2 \right ]$

Option 3)
$\left [-2, 4 \right ]$

Option 4)
None of these

Answers (1)

best_answer

Results from General Solution -

If $\sin ^{2}\Theta = \sin ^{2}\alpha$

$\cos ^{2}\Theta = \cos ^{2}\alpha$

$\tan ^{2}\Theta = \tan ^{2}\alpha$

then $\Theta = n\pi \pm \alpha$

- wherein

$\alpha$ is the given angle

$x= \sin ^{2}\alpha -\cos ^{2}\alpha$

$\Rightarrow 2\sin ^{2}\alpha - (1-2\sin ^{2}\alpha )$

$\Rightarrow 2\sin ^{2}\alpha -1+2\sin ^{2}\alpha$

$\Rightarrow 4\sin ^{2}\alpha -1+2\sin ^{2}\alpha$

$4\sin ^{2}\alpha -1$ $\therefore 0\leq \sin ^{2}\alpha \leq 1$

$0-1, 4-1$

$\left [ -1,3 \right ]$

Option 1)

$\left [-1, 3 \right ]$

This solution is correct.

Option 2)

$\left [1, 2 \right ]$

This solution is incorrect.

Option 3)

$\left [-2, 4 \right ]$

This solution is incorrect.

Option 4)

None of these

This solution is incorrect.

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